When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 12.0 fluid ounce drink from 75°F to 35°F, if the heat capacity of the drink is 4.18 J/g·°C? (Assume that the heat transfer is 100% efficient.)

please explain realy realy well

thanks:)

this is wut i got:

12oz=340.194 grams

(miX330j/g) + [(340.194gX4.184X40]=0

56934.86 + 330j/g(mi)=0

-56934.86=330j/g(mi)

-56934.86/330j/g = -172.5298=(mi)

Here is what I posted previously.

Convert 12 ounces to grams.
(mass ice x 330 J/g) + [(mass water x specific heat water x (Tfinal-Tinitial)] = 0
Solve for mass ice.

Show your work and I'll try to find the error. Mass ice is the only unknown. mass water is 12 ounces; you need to convert that to grams. Specific heat water you know as well as Tf and Ti. There is no explaining to do.

oops sorry I am rebekah

my sis changed it 4 her

ok. The error is in the 40. It should be a -40. It works this way.

(mass ice x 330) + [(mass water x specific heat water x (Tfinal-Tinitial)] = 0

(330X) + [(340.2 x 4.184 x (35-75)] = 0
330X + [(340.2 x 4.184 x (-40)] = 0
I will let you finish but I think you can see that it changes only the sign so you end up with +172 g and not -172 g(the negative sign makes no sense, of course).

Some quick comments. The F must be converted to C, the ice, when it melts at zero C must also be heated to 35 F (converted to C), and the fluid ounce is not the same as the the ounce in mass units.

how much ice is required to cool a 10.0 oz drink from 80 f to 31 f, if the heat capacity of the drink is 4.18 j/gc?

This answer is fine because -40 F is equal to -40 C

I got the answer 172 g of ice needed.

To solve this problem, we need to calculate the amount of heat energy that needs to be absorbed by the ice to cool the drink. Then, we can determine the quantity of ice required using the energy absorbed per gram of ice.

Step 1: Convert the volume of the drink from fluid ounces to grams.
Since the density of water is 1 g/mL, we can assume that the density of the drink is also 1 g/mL. Thus, 12.0 fluid ounces of the drink is equal to 12.0 ounces or 12.0 x 28.35 grams (since 1 ounce = 28.35 grams) which equals 340.2 grams.

Step 2: Calculate the temperature change of the drink.
The temperature change is the final temperature minus the initial temperature.
ΔT = 35°F - 75°F = -40°F

Step 3: Convert the temperature change to Celsius.
ΔT = -40°F x (5/9) = -22.22°C (rounded to two decimal places)

Step 4: Determine the amount of heat energy required to cool the drink.
The heat energy can be calculated using the formula:
Heat energy (Q) = Mass (m) x Specific heat capacity (c) x ΔT

Q = 340.2 grams x 4.18 J/g·°C x -22.22°C
Q = -32,410 J

Step 5: Convert the heat energy to kilojoules.
Since the heat energy per gram absorbed by ice is given as 0.33 kJ, we can calculate how much ice is required by dividing the total heat energy by the heat energy absorbed per gram.
Ice (m) = Q / heat energy per gram
Ice (m) = -32,410 J / 0.33 kJ (since 1 kJ = 1000 J)
Ice (m) = -98.51 grams

Since mass cannot be negative, the negative sign indicates that the heat energy is released, not absorbed, by the ice. Therefore, to cool the drink from 75°F to 35°F, approximately 98.51 grams of ice are required.